Let A be a point on the line

Let point $\mathrm{A}$ is

$(1-3 \mu) \hat{\mathrm{i}}+(\mu-1) \hat{\mathrm{j}}+(2+5 \mu) \hat{\mathrm{k}}$

and point $B$ is $(3,2,6)$

then $\overrightarrow{\mathrm{AB}}=(2+3 \mu) \hat{\mathrm{i}}+(3-\mu) \hat{\mathrm{j}}+(4-5 \mu) \hat{\mathrm{k}}$

which is parallel to the plane $x-4 y+3 z=1$

$\therefore 2+3 \mu-12+4 \mu+12-15 \mu=0$

$8 \mu=2$

$\mu=\frac{1}{4}$